Question

Why does ionic stuff dissolve?

Answer 1

That is, why does being surrounded by polar water molecules correspond to less potential energy than the ionic lattice not interactingwith the water? Surely more energy is used up in breaking the strong ionic bonds than is released forming the partially ionic bonds with water (as they have only a small charge, after all).

Answer 2

*?

Answer 3

it first depends on the lattice cause im sure you know that there are compounds which are hard to dissolve in water... but the hidratation energy is often big enough to brake the surface bonds of compounds since they are not in equilibrium like the atoms inside the crystal. need more?

Answer 4

WHY is it often big enough is the essence of my question.

Secondly (because I have to go soon): if such a stable arrangement, why don't things (even when not dissolved) bond in a similar way to the water-ion system when dissolved (with regards to the shape of the ionic bonds).
I will return in an hour and a half, but don't worry- I WILL thoroughly read through your answer then. Sorry I have to leave in the first place.

Answer 5

i suggest you to read this since you have to go soon, put a bookmark on this page and read later, there you got even a table of hydration energy for some ions:
http://www.science.uwaterloo.ca/~cchieh/cact/applychem/hydration.html

Answer 6

All dissoluations are driven by the tremendous increase in entropy that accompany them. It is exceedingly rare for the interactions between solute and solvent to be stronger than those between the solute particles, and between the solvent particles, both of which are disrupted by the dissolution. That is, usually dissolution costs energy -- is enthalpically uphill. However, the increase in entropy more than compensates, at a high enough temperature.

If you are asking how it happens at the mechanical level, not the thermodynamic level, then the answer is that, roughly speaking, from time to time you get a surge in the local energy available, e.g. a wave of solvent molecules crash onto the ionic solid with unusual force, and break off a few ions. This situation is not energetically stable -- if the ions were to reconnected to the crystal matrix, they would gladly re-insert into the crystal lattice and release energy.

But they don't get the chance, because as soon as they are freed from the lattice, their translational energy takes them far away -- remember, they are moving at several hundred meters per second! Ordinarily, this translates to just a furious vibration in place, in the crystal lattice, because their motion is contrained by the powerful electrostatic forces from neighboring ions. But for a moment, these forces have been removed, because the ions have been detached from the lattice (and surrounded by water molecules). So the ions quite rapidly drift away, and in a few femtoseconds are too far away from the lattice to feel the electrostatic forces. They continue drifting, and, since they have the entire volume of the solution in which to drift, and the original crystal lattice is only a very small part of that volume, the chances are just very low that they will stumble back across the crystal lattice before the lattice has entirely dissolved.

Once the lattice has entirely dissolved, the problem of re-integration becomes even more problematic, because now you have the issue that two or three ions right next to each other is not especially stable. To recreate the stable crystal lattice you have to have many ions, right next to each other in exactly the right geometrical arrangement, with exactly the right number of cations and anions, et cetera. If this arrangement could form, then it would be energetically favored -- more stable than the dissolved ions. But it's just too unlikely, for all those ions to come together in exactly the right way, all at most. So it doesn't happen. This is the microscopic meaning of the entropy loss on recrystallization being too high.

Now, when the temperature is lowered, things move slower, and with ions moving much more slowly the forces between them have more time to jiggle them into the right places to reform the lattice, so it becomes more likely. At some point, it becomes likely enough that it happens in the time you have to observe the experiment.

It may be worth noting that if you were willing to observe your solution with dissolved crystal long enough -- say, 10^100 times the age of the Universe -- you certainly would, sooner or later, see the crystal re-assemble itself. But in the ordinary way, over a period of minutes, hours or days, it's far too unlikely to observe.

Answer 7

Up until now I have used as a rule of thumb in grokking why chemical reactions happen that potential energy always decreases (that is, energy becomes more spread out). Is entropy, then, a better tool to use, given that decrease of energy is not the only form of entropy increase, so one can have entropy increase with energy spreading out?

Thanks, you have unwittingly answered another of my questions that I had forgotten about concerning why temperature affects the amount of solid that can dissolve with temperature.

On a related note, why does gas do the reverse (become less soluble with increasing temperatures)?

Answer 8

You can't use energy or entropy exclusively, because it's a balance between the two, except at T=0 (when only energy matters) or T=infinity (when only entropy matters). That's one good reason for the existence of the concept of free energy: you can be confident that whenever the free energy decreases, the process is spontaneous.

Part of the problem is that you have to think out very carefully the nature of your system and the restrictions on it. Most realistic systems have some, but not completely free, interactions with the environment, and that complicates matters. This is where the brilliant exposition of thermodynamics by people like Boltzmann and Gibbs becomes so valuable, because their clear thinking allows you to understand the criteria for spontaneity.

If you have an *isolated* system -- no contact with the environment -- then its energy is conserved, and cannot change, and the only spontaneous processes are those that increase the entropy. If you allow the system to exchange energy via heat with the surroundings, e.g. hold it at a constant temperature, then processes that lower the Helmholtz free energy (the Legendre transform of the entropy with respect to energy) are spontaneous. If you also allow it to exchange volume with the surroundings, e.g. hold it at a constant pressure, then processes that lower the Gibbs free energy (Legendre transform of the entropy with respect to energy and volume) are spontaneous.

Answer 9

Gases dissolving is the opposite of solids dissolving. The gas is going from being free to move around in a large volume to being confined in a smaller volume (the liquid solution). So entropy *decreases*. Dissolution of gases is enthalpically driven: the gas particles ordinarily have zero interactions with each other, and when they dissolve they have some weak attractive intermolecular interactions with the solvent. So the enthalpy of a gas does indeed go down when it dissolves. This is just the opposite of a solid dissolving, where the enthalpy goes up and the entropy goes down.

To see the change with T, just use the usual connection between free energy, enthalpy and entropy:

dG = dH - T dS

Remember also, vide supra, dG < 0 means the process is spontaneous.

Now for solids dissolving in liquids, dH > 0 and dS > 0, so:

dG < 0 at high T, dG > 0 at low T.

We conclude solids dissolve more readily at higher temperatures.

For gases dissolving in liquids, dH < 0 and dS < 0, so:

dG > 0 at high T, dG < 0 at low T.

We conclude gases dissolve more readily at low temperatures.

Answer 10

Incidentally, one of the reason you may be used to thinking in terms of energy reduction is you have learned some physics. It's worth keeping in mind that physics is almost all done implicitly at T=0. For example, all mechanics assumes that the energy per moving degree of freedom is hugely more than kT -- which means that, effectively, the moving system might as well be at T=0.

kT is about 10^-21 joules per degree of freedom at room temperature. So if you are considering two 500g objects on a frictionless track colliding at 1m/s, the energy in their center of mass motion is about 10^22 times the energy available per degree of freedom in the matter surrounding them. You could lower the temperature of the surroundings to T=0 without affecting things in the slightest. That's what I mean by saying that mechanics is essentially always implicitly done at T=0. Under those conditions, energy dominates.

But when it comes to masses the size of atoms and molecules colliding, it is no longer true that the energy available fantstically exceeds what's avaiable in the environment, so you are no longer at T=0, and you cannot use the simplifying assumptions you use in mechanics class.